Problem: $\overline{AC}=10$ $\overline{AB} = {?}$ $A$ $C$ $B$ $?$ $10$ $ \sin( \angle BAC ) = \frac{7\sqrt{149} }{149}, \cos( \angle BAC ) = \frac{10\sqrt{149} }{149}, \tan( \angle BAC ) = \dfrac{7}{10}$
Solution: $\overline{AC}$ is adjacent to $\angle BAC$ $\overline{AB}$ is the hypotenuse (note that it is opposite the right angle) SOH CAH TOA We know the adjacent side and need to solve for the hypotenuse so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}} = \frac{10}{\overline{AB}} $ $ \overline{AB}=\frac{10}{\cos( \angle BAC )} = \frac{10}{\frac{10\sqrt{149} }{149}} = \sqrt{149}$